Stability constants of complexes — chelate effect: Visual walkthrough

Step 0 — The one piece of notation we must earn first: square brackets

WHAT. Chemists wrap a whole metal-plus-attached-ligands unit in square brackets to say "this is one coordination complex, a single dissolved species". So $[M (H_{2} O)_{6}]$ means "one metal ion $M$ carrying six water molecules on its spokes" — read the whole bracket as one object.

WHY define it now. Every equation below uses this shorthand. If we don't earn it here, the brackets in Step 4 would be a symbol used before it was defined.

Step 1 — What "stable" even means: a tug-of-war in water

WHAT. A metal ion in water is never bare. Six water molecules already cling to it. Adding a ligand $L$ is a swap: $L$ elbows a water out and takes its place. We are watching a competition, not a first grab.

WHY start here. If we forget the waters, the whole entropy story disappears. Every count we make later is a count of free particles floating in solution, so we must first see who is bound and who is free.

PICTURE. The central amber circle is the metal $M$ . Six white spokes end in cyan water molecules — the crowd already holding on. To bind a new ligand, a water must be released into the free solution (the loose molecules drifting at the edges).

Figure — Stability constants of complexes — chelate effect

Step 2 — One tooth vs two teeth: what "denticity" looks like

WHAT. Two ligand shapes enter the ring. Ammonia ( $NH_{3}$ ) has one donor nitrogen — it grips a single spoke: it is monodentate (one tooth). Ethylenediamine ("en", $H_{2} N -CH_{2} -CH_{2} -NH_{2}$ ) has two nitrogens joined by a short carbon bridge — it grips two spokes at once: it is bidentate (two teeth).

WHY it matters. Both offer the same kind of donor atom (nitrogen). So the main difference in stability cannot come from the atom type — it comes largely from how many free particles each releases (with a smaller ring/strain contribution we handle in Step 7). That is the fair fight we set up now. (See Denticity and ligand classification.)

PICTURE. Left: a lone $NH_{3}$ , one dot, reaches for one spoke. Right: "en" is a hinged clamp — two dots on one backbone — reaching for two spokes. The bridge is the string that ties the two teeth together.

Step 3 — Turn a complex into a number: the stability constant

WHAT. We need a single number that says how far the swap goes. That number is the formation (stability) constant. For the first ligand added:

$K_{1} = products over reactants \frac{[ M L ]}{[ M ] [ L ]}$

$[M L]$ — concentration of the formed complex (top: what we made).
$[M]$ — concentration of free metal still uncomplexed (bottom).
$[L]$ — concentration of free ligand still floating (bottom).
Big $K_{1}$ ⇒ the top wins ⇒ lots of complex ⇒ stable.

(Here the square brackets mean concentration of — the second use we flagged in Step 0.)

To build the full $M L_{6}$ we chain six such steps; multiplying them gives the overall constant $β_{6} = K_{1} K_{2} \dots K_{6}$ , and its logarithm $lo g β$ is the score we compare.

WHY a logarithm. Because the constants multiply over steps, their logs add — turning a runaway product into a friendly sum we can read off a bar chart. (Link: Gibbs free energy and equilibrium constant.)

PICTURE. A number line of $lo g β$ : ammonia's six-step complex sits low; en's three-step complex sits far to the right. The gap between them is the entire mystery of this page.

Step 4 — Count the free particles: the bookkeeping that decides everything

WHAT. Write both swaps with the spectator waters shown, then count independent dissolved species on each side (Step 0: each complex ion counts as one, each free molecule counts as one).

Ammonia route — six separate teeth: $[M (H_{2} O)_{6}] + 6 N H_{3} ⇌ [M (N H_{3})_{6}] + 6 H_{2} O$ Free particles: $1 + 6 = 7$ on the left, $1 + 6 = 7$ on the right. No change: $7 \to 7$ .

En route — three double-teeth: $[M (H_{2} O)_{6}] + 3 e n ⇌ [M (e n)_{3}] + 6 H_{2} O$ Free particles: $1 + 3 = 4$ on the left, $1 + 6 = 7$ on the right. Gain of three: $4 \to 7$ .

WHY this is the crux. Nature counts free particles. En consumes 3 molecules but releases 6 waters — a net +3 loose particles set wandering. Ammonia consumes 6 and releases 6 — it breaks even. That surplus of freedom is the chelate effect in raw form.

PICTURE. Two balance-beam scales. Top (ammonia): equal blobs both sides — level beam, "7 = 7". Bottom (en): fewer blobs consumed, more waters spilled out — the right pan overflows with three extra free waters, tilting toward more disorder.

Step 5 — Turn "extra free particles" into energy: entropy and $Δ G$

WHAT. More free particles = more ways to arrange the solution = higher entropy $S$ (a measure of disorder — see Entropy and the second law). The master equation ties disorder to whether a reaction runs:

$Δ G^{\circ} = bond energy change Δ H^{\circ} - freedom \times temperature T Δ S^{\circ}$

$Δ G^{\circ}$ — the "will it go?" number. More negative ⇒ more product ⇒ bigger $β$ .
$Δ H^{\circ}$ — heat from making/breaking bonds. Because both routes make the same kind of M–N bond, this is close for the two — but not exactly equal (see the note below).
$Δ S^{\circ}$ — the disorder change. Positive and bigger for en (three extra free waters).
$T$ — absolute temperature; $- T Δ S^{\circ}$ becomes more negative when $Δ S^{\circ}$ is positive.

The $- T Δ S^{\circ}$ term is the main referee — and it favours en — but the enthalpy is not a perfect tie:

WHY these tools. We needed a bridge from "particle counting" (Step 4) to "which complex forms". $Δ S^{\circ}$ is that bridge (it counts arrangements), and $Δ G^{\circ} = Δ H - T Δ S$ converts disorder into a yes/no on the reaction.

PICTURE. Two energy bars. $Δ H^{\circ}$ (cyan) is nearly the same height for both (a small extra dip for en shown hatched). The amber $- T Δ S^{\circ}$ block is short for ammonia but tall for en, pushing en's total $Δ G^{\circ}$ far deeper into the negative "downhill" region.

Step 6 — From energy back to the score: how big is the jump?

WHAT. Connect $Δ G^{\circ}$ to the number we started with, $lo g β$ :

$lo g β = \frac{- Δ G ^{\circ}}{2.303 R T}$

$R = 8.314 J K^{- 1} mol^{- 1}$ — the gas constant (energy per degree per mole).
$T = 298 K$ — room temperature.
$2.303$ — the factor converting natural log to base-10 log.

Where does $Δ S^{\circ} \approx + 84 J K^{- 1} mol^{- 1}$ come from? It is an estimate, not a bare assertion. The net change of Step 4 is +3 independent particles in solution. In liquid water, the entropy "cost" of tying up one small solute (or the "gain" of releasing one) is empirically around $+ 25$ to $+ 30 J K^{- 1} mol^{- 1}$ per particle — this is the translational/rotational freedom a released water molecule regains once it stops being locked on a spoke. Three released particles then give roughly $Δ S^{\circ} \approx 3 \times 28 \approx + 84 J K^{- 1} mol^{- 1} .$ Measured chelate reactions land in exactly this ballpark, which is why we use it. The extra downhill push is: $- T Δ S^{\circ} = - (298) (84) = - 2.50 \times 1 0^{4} J/mol = - 25 kJ/mol$ Convert to score units: $Δ (lo g β) = \frac{25000}{2.303 ( 8.314 ) ( 298 )} \approx 4.4$

WHY this closes the loop — honestly. Entropy alone buys roughly +4.4 in $lo g β$ . The observed gap between $[M (N H_{3})_{6}]$ and $[M (e n)_{3}]$ is larger, about +9.7. So entropy accounts for about half of the gap; the remainder comes from the enthalpy contributions of Step 5 (favourable 5-ring geometry, plus per-step statistical and electrostatic factors that make en's later binding steps stronger than ammonia's). The lesson: entropy is the biggest single reason and the one people forget — but it is not the whole 9.7. Our estimate predicts the right order of magnitude of the effect, and points to where the rest lives.

PICTURE. A conversion pipeline lifting the en bar by +4.4 from entropy, with a second amber block labelled "+ enthalpy / geometry" carrying it the rest of the way up to the observed +9.7.

Step 7 — The edge cases: when the effect shrinks or dies

WHAT. The chelate effect is not unconditional. Its size depends on the ring the ligand folds into, and on any pre-shaping. Cases:

5-membered ring (en + M): maximum. Bite angle is unstrained; both teeth reach comfortably — and this even lends a small favourable ΔH°. ✅
4-membered ring: strained. Teeth too close; forcing the metal into a tight ring costs bond energy ( $Δ H^{\circ}$ turns unfavourable). Effect weakens. ⚠️
Very large ring: floppy. The second tooth rarely finds the metal — low effective concentration, ring often fails to close. Effect fades. ⚠️
Zero teeth (monodentate limit): no effect at all. With one tooth per molecule there is no net particle gain — this is the ammonia baseline. ➖
Pre-organised macrocycle (crown/cryptand): even bigger — the macrocyclic effect. The ring is already the right shape, so it pays no penalty to wrap up, improving both ΔS° and ΔH°. ✅✅ (See Macrocyclic and cryptand ligands.)

WHY show all five. A rule you can't break isn't a rule you understand. Seeing where the effect peaks (5-ring), degrades (4 or huge), vanishes (mono), or super-charges (macrocycle) proves it is really about geometry enabling the particle-count gain — not magic. It also shows where the leftover enthalpy contribution of Step 6 comes from.

PICTURE. A stability-vs-ring-size curve: rising to a sharp peak at 5, dropping on both sides; a lone dot far above the curve marks the pre-organised macrocycle; a flat baseline marks the monodentate "no effect" case.

The one-picture summary

Everything on one canvas: two teeth tied by a string drop into the metal, kicking out six waters while only three ligands go in — three extra free particles spill into solution → entropy rises (≈ +4.4 in $lo g β$ ) → then favourable ring geometry (enthalpy) carries it the rest of the way → the observed leap of about +9.7. The whole derivation, one frame.

Recall Feynman retelling of the whole walkthrough

Picture the metal as a kid clutching six water balloons. You want your balloons on that kid instead. (First, one bit of shorthand: whenever I draw a box around "kid + balloons", that whole thing counts as one floating object in the room — no matter how many balloons are inside.)

Plan A (ammonia): hand over six separate balloons. The kid drops six, grabs six. The room has exactly as many loose objects as before — nobody got freer, nothing exciting happened. Small $lo g β$ .

Plan B (en): hand over three balloons tied in pairs by string. The kid drops six waters but only takes three tied bundles. Now three extra objects float loose in the room — the room got livelier, more disordered, more free. Nature adores extra freedom (that's entropy), so it rewards this trade — worth about +4.4 on the scoreboard.

The honest total: the observed reward is bigger, about +9.7. Roughly half is that freedom (entropy); the other half comes because the tied bundle also happens to loop around the kid at a comfortable angle (a 5-atom loop), which makes the grip itself a touch better (enthalpy). So it's mostly freedom, plus a real geometry bonus — not stronger glue alone.

The catch: the string must be just the right length — a 5-atom loop. Too short (4) and the bundle is awkwardly tight; too long and the second balloon dangles and never gets grabbed. And if the string is pre-tied into a perfect loop (a macrocycle), the reward is bigger still.

Recall

Why is $[N i (e n)_{3}]^{2 +}$ far more stable than $[N i (N H_{3})_{6}]^{2 +}$ despite identical N-donors? ::: Mainly because 3 en release 6 waters while consuming only 3 molecules (4→7 free particles), giving a large positive $Δ S^{\circ}$ ; this accounts for about half the $lo g β$ gap, with favourable 5-ring geometry (enthalpy) supplying the rest. In the swap $[M (H_{2} O)_{6}] + 6 N H_{3} \to [M (N H_{3})_{6}] + 6 H_{2} O$ , what is the net change in free particles? ::: Zero — 7 free particles on each side, so no entropy driving force. Does entropy alone explain the whole ~9.7 gap in $lo g β$ ? ::: No — entropy buys ≈ +4.4 (about half); the remaining ≈ +5 comes from favourable ring/enthalpy and per-step statistical/electrostatic factors. What size chelate ring is most stable and why? ::: 5-membered — the unstrained bite angle gives no enthalpy penalty (even a small bonus), letting the entropy gain dominate.